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exponential_function [2015/12/09 15:10] nikolaj |
exponential_function [2016/07/10 15:08] nikolaj |
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^ $\mathrm{e}^z \neq 0 $ ^ | ^ $\mathrm{e}^z \neq 0 $ ^ | ||
- | ^ $\frac{\mathrm d}{\mathrm d z}\mathrm{e}^{f(z)} = \frac{\mathrm d}{\mathrm d z}f(z)\cdot \mathrm{e}^{f(z)} $ ^ | + | ^ $\frac{\mathrm d}{\mathrm d z}\mathrm{e}^{f(z)} = \frac{\mathrm d}{\mathrm dz}f(z)\cdot \mathrm{e}^{f(z)} $ ^ |
$a,b,r,\theta\in\mathbb R$ | $a,b,r,\theta\in\mathbb R$ | ||
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so | so | ||
- | $\left(1 + \dfrac {x} {n} \right)^n = \sum_{k=0}^n \left( \dfrac {n!} {(n-k)!\,n^k} \right) \dfrac {x^k} {k!} $ | + | $\left(1 + b(n)\,x \right)^n = \sum_{k=0}^n \left( b(n)^{-k}\dfrac {n!} {(n-k)!} \right) \dfrac {x^k} {k!}$ |
- | $= \sum_{k=0}^n \dfrac {x^k} {k!}\cdot\prod_{j=1}^{k}\left(1-\dfrac{k-j}{n}\right)$ | + | (Note that here the summands depend on the upper sum bound $n$, this sum doesn't make for an infinite sum of partial sums - the to be partial sums are all different) |
+ | |||
+ | So | ||
+ | |||
+ | $\left(1 + \dfrac {x} {n} \right)^n = \sum_{k=0}^n \left( \dfrac {n!} {(n-k)!\,n^k} \right) \dfrac {x^k} {k!} = \sum_{k=0}^n a_k(n)\dfrac {x^k} {k!}$ | ||
+ | |||
+ | with | ||
+ | $a_k(n)=\prod_{j=1}^{k}\left(1-\dfrac{k-j}{n}\right)$ | ||
+ | |||
+ | also | ||
$= \sum_{k=0}^n \prod_{j=1}^{k}\left(\dfrac{1}{j}-\dfrac{1}{n}\left(\frac{k}{j}-1\right)\right)x$ | $= \sum_{k=0}^n \prod_{j=1}^{k}\left(\dfrac{1}{j}-\dfrac{1}{n}\left(\frac{k}{j}-1\right)\right)x$ |